Question

$\frac{{1.2}^2+{2.3}^2+{3.4}^2+.\cdot .\cdot .\cdot .\cdot +n(n+1{)}^2}{1^2.2+2^2.3+3^2.4++n^2(n+1)}=$

• A. $\frac{3n+1}{3n+5}$
• B. $\frac{3n+5}{3n+1}$
• C. $(3n+1)(3n+5)$
• D. $\frac{3n+5}{3n+7}$

Answer 1

The correct option is B $\frac{3n+5}{3n+1}$

Numerator : ${\sum }_{k=1}^{n}k(k+1{)}^2$ $={\sum }_{k=1}^n(k^3+2k^2+k)$
Denominator : ${\sum }_{k=1}^n{k}^2(k+1)$ = ${\sum }_{k=1}^n(k^3+k^2)$
On evaluating each sum we get
Numerator = $n(n+1)(\frac{n(n+1)}{4}+\frac{2n+1}{3}+\frac{1}{2})$
Denominator = $n(n+1)(\frac{n(n+1)}{4}+\frac{2n+1}{6})$
Now,
$Answer=\frac{numerator}{denominator}=\frac{n(n+1)}{n(n+1)}\times \frac{(n+2)(3n+5)}{(n+2)(3n+1)}=\frac{3n+5}{3n+1}$
Hence, option 'B' is correct.